98. Validate Binary Search Tree

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本文探讨了如何验证一个二叉树是否为有效的二叉搜索树。通过两种方法实现：一是递归地检查每个节点的值是否在其合法范围内；二是进行中序遍历并将节点值保存到数组中，然后检查数组元素是否递增。

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98. Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.

Example 1:

2
/ \
1 3

Binary tree [2,1,3], return true.

Example 2:

1
/ \
2 3

Binary tree [1,2,3], return false.

【解法一】我的解法，引入最大最小值

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        return isvalid(root, LONG_MIN, LONG_MAX);
    }
    bool isvalid(TreeNode *root, long MIN, long MAX) {
        if (!root) return true;
        if (root->val <= MIN || root->val >= MAX) return false;
        return isvalid(root->left, MIN, root->val) && isvalid(root->right, root->val, MAX);
    }
};

【解法二】网上的其他解法,中序遍历存入数组中，判断a[i]>a[i+1]

class Solution {
public:
    vector<int> temp;
    bool isValidBST(TreeNode* root) {
        inOrderTraversal(root);
        if(temp.size()<=1) return true;
        for(int i=1;i<temp.size();i++){
            if(temp[i]<=temp[i-1]) return false;
        }
        return true;
    }
    void inOrderTraversal(TreeNode * root){
        if(!root) return;
        inOrderTraversal(root->left);
        temp.push_back(root->val);
        inOrderTraversal(root->right);
    }
};